Question

LIMITS-Seems easy but somehow I don't know what to do with the fractions
Find the limit(if it exists) as x approaches 0 for the function (1/(X+1)-1)/x .

Answer 1

\[\frac{ 1 }{ \frac{ (x+1)-1 }{ x } }\]
Something like that? Can't quite tell how you meant to type it up/

Answer 2

Oops let me try again. ((1/(x+1))-1)/x

Answer 3

\[\frac{ 1 }{ x+1 }-\frac{ 1 }{ x }\] correct now?

Answer 4

Oh, I think I know what you meant to type up:

\[\frac{ \frac{ 1 }{ x+1 } -1}{ x }\]

Answer 5

no
the first fraction is subtrated by 1, then that whole quantity is divided by x.
oh yeah thats it c:

Answer 6

Okay, so just making sure we do our algebra right then. Alright, so this is how Id do it. Id make that top portion all into one fraction by making a common denominator. Basically, id take that -1 and multiply top and bottom by x+1, which allows me to make it into one fraction like this:

\[\frac{ \frac{ 1-(x+1) }{ x+1 } }{ x }\implies \frac{ \frac{ -x }{ x+1 } }{ x }\]
Now this is just the division of two fractions. The first fraction is -x/(x+1) and the second fraction is x/1. So I can rewrite what we have like this:

\[\frac{ -x }{ x+1 }\div \frac{ x }{ 1 }\implies \frac{ -x }{ x+1 }*\frac{ 1 }{ x }=\frac{ -x }{ x(x+1) }\]

So of course the x on tiop and bottom cancel out to leave
\[\frac{ -1 }{ x+1 } \]Now apply your limit : )

Answer 7

Aha! I can see it clearly now. The limit is -1.

Answer 8

yep yep :3 Honestly, its often not the calculus that sucks, its the algebra, lol.

Answer 9

Haha I can see it coming back to haunt me already. Anyway that was superb, thanks for your time. :D

Answer 10

No problem : )